Convergence of Taylor Series of Function Analytic on Disk/Lemma

From ProofWiki
Jump to navigation Jump to search

Lemma

Let $y > 1$.


Then:

$\ds \lim_{n \mathop \to \infty} \frac n {y^n} = 0$


Proof

Note that $\ln y > 0$ as $y > 1$.

\(\ds \lim_{n \mathop \to \infty} \frac n {y^n}\) \(=\) \(\ds \lim_{n \mathop \to \infty} \frac n {\paren {e^{\ln y} }^n}\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \frac n {e^{\paren {\ln y} n} }\)
\(\ds \) \(=\) \(\ds 0\) as $\ds \lim_{x \mathop \to \infty} \frac x {e^{\paren {\ln y} x} } = 0$ by Limit at Infinity of Polynomial over Complex Exponential as $\ln y > 0$

$\blacksquare$